Question

A particle of mass $2\, kg$ moves on a smooth horizontal plane under the action of a single force $\vec{F}=(3 \hat{i}+4 \hat{j}) N$. Under this force it is displaced from $(0,0)$ to $(1 \, m , 1 \, m )$. The initial speed of the particle is $\sqrt{3} m / s$ and final speed is $\sqrt{5 k} \, m / s$. Find the value of $k$.

Answer 1

$W=\Delta K E=\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}$
Here, $W=\vec{F} \cdot \vec{d}$
$=(3 \hat{i}+4 \hat{j}) \cdot(\hat{i}+\hat{j})$
$=3+4=7\, J$
$\Rightarrow 7=\frac{1}{2} \times 2 \times v_{2}^{2}-\frac{1}{2} \times 2 \times(\sqrt{3})^{2}$
$\Rightarrow v_{2}^{2}=10$
$ \Rightarrow v_{2}=\sqrt{10} \,m / s$